Vertical Line through z=0 in complex plane mapped with f(z)=(1+z)/(1-z)

Question

I have the vague notion that the imaginary axis maps to a circle with f(z)=(1+z)/(1-z). $$ \begin{array}{lll} f(\infty) & = & -1\\ f(i) & = & e^{\pi/4}\\ f(-i) & = & e^{-\pi/4}\\ f(0) & = & 1\\ \end{array} $$ Seems to be mapping only in the right side of a unitary circle. Is this correct?

Answer 1

No. It maps the imaginary axis onto the unit circle minus $-1$. Note that, if $t\in\mathbb R$,$$\frac{1+it}{1-it}=\frac{(1+it)^2}{(1+it)(1-it)}=\frac{1-t^2}{1+t^2}+\frac{2t}{1+t^2}i$$and the complex numbers of this form are precisely the elements of the unit circle (with one exception: $-1$).